Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $r = \dfrac{y - 1}{y^2 - 100} \div \dfrac{y^2 + 6y}{y^3 - 4y^2 - 60y} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{y - 1}{y^2 - 100} \times \dfrac{y^3 - 4y^2 - 60y}{y^2 + 6y} $ First factor out any common factors. $r = \dfrac{y - 1}{y^2 - 100} \times \dfrac{y(y^2 - 4y - 60)}{y(y + 6)} $ Then factor the quadratic expressions. $r = \dfrac {y - 1} {(y - 10)(y + 10)} \times \dfrac {y(y - 10)(y + 6)} {y(y + 6)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {(y - 1) \times y(y - 10)(y + 6) } { (y - 10)(y + 10) \times y(y + 6)} $ $r = \dfrac {y(y - 10)(y + 6)(y - 1)} {y(y - 10)(y + 10)(y + 6)} $ Notice that $(y - 10)$ and $(y + 6)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {y\cancel{(y - 10)}(y + 6)(y - 1)} {y\cancel{(y - 10)}(y + 10)(y + 6)} $ We are dividing by $y - 10$ , so $y - 10 \neq 0$ Therefore, $y \neq 10$ $r = \dfrac {y\cancel{(y - 10)}\cancel{(y + 6)}(y - 1)} {y\cancel{(y - 10)}(y + 10)\cancel{(y + 6)}} $ We are dividing by $y + 6$ , so $y + 6 \neq 0$ Therefore, $y \neq -6$ $r = \dfrac {y(y - 1)} {y(y + 10)} $ $ r = \dfrac{y - 1}{y + 10}; y \neq 10; y \neq -6 $